An experiment is a procedure that gives a set of outcomes. This set of all possible outcomes is called the Sample Space, S
We will for now, assume that 0<∣S∣<+∞
We also make the basic assumption that each outcome in the sample space, S, is equally likely to happen
An event, E, is a subset of S - That is E⊆S
Definition The probability of an event E⊆S, where S is a finite sample space with equally likely outcomes is
P(E)=∣S∣∣E∣
A typical simple sample space S would be the roll of a fair die, or the toss of a pair of dice, the cut of a deck of playing cards, the choice of 6 numbers ont of 49 etc
Example 1
Consider a 6 sided fair die
The outcomes possible are
S1={1,2,3,4,5,6}
Let E be the event that the outcome is an even integer
Then E={3,4,6} and P(E)=63=21.
Example 2
Consider the toss of two fair distinguishable dice
Then the sample space is, using the same notation as Exercise 1,
Exercise What is the minimum number, n, such that P(E)⩾21 ? Hint : this implies P(Eˉ)<21 This is best done by trial and error?
Let's Make a Deal
You are given the choice of picking one of 3 doors to win the grand prize
You pick a door. The host then opens one of the other 2 doors and shows you that it was a "losing door", You are then offered the choice of switching doors
What should you do?
Since the doors are equally likely that they hide the jackpot, the probability that your choice was correct is 31, and wong is 32.
Thu the probability that the other two dooms hid the jackpot is 32, and since this is not affected by the opening of a losing dore, the probability that your door is wrong is still 32 ?
Thus the other door is the night choice with a probability of 32, so switch doors
What in the probability that a card selected from a deck is an ace?
Solution
Let S={ deck of cards, ∣S∣=52
Let E={ an ace is picked },∣E∣=4
∴P(E)=∣S∣∣E∣=524=131
Example 4
What is the probability that a 5 card poker hand contains the ace of hearts
What is the probability that a 5 card poker hand has two pairs?
Solution Let E={5 card hands with two pairs }
Then ∣E∣=pick the value of the pairs→(132)⋅(42)⋅(42)⋅(441)→pick the fifth card↑pick the two cards with one of the given values=2⋅113⋅12⋅2⋅14⋅3⋅2⋅14⋅3⋅44=13⋅63⋅44
So far we have dealt only with the very restricted case where our sample space S is finite and each outcome in (ie., element of) S and is equally likely
This is insufficient on two counts
We often deal with experiment where the sample space is not finite. when S is infinite it can be countable or uncountable
When S is countable ∣S∣≤ℵ0 we talk about Discrete sample spaces and when S is uncountable we talk about Continuous sample spaces The analysis of Continuous Probability requires calculus methods (see M2236)
All outcomes need not he equally, likely
For example an "unfair" die 1 may come up more often than 6 does
To deal with these situations we will develop a theory of Probability for Discrete Sample Spaces
Let S be a countable sample space (finite or infinite) and s∈S be an outcome. We assign a probability p(s) to each outcome. so that P:S→R satisfies the following
i) 0⩽p(s)≤1
ii) s∈S∑p(s)=1
Note i) ⇒{no negative probabilities, no probabilities >1
ii) If ∣S∣=n<+∞, then with S={s1,⋯,sn},
k=1∑np(sk)=1
If ∣S∣=ℵ0, then with S={s1,s2,…}
k=1∑∞p(sk)=1convergent infinite series
This imposes "severe" restrictions on p
If p:S→R satisfies i) and ii) we call p(s) a probability distribution on S
Now let E⊆S be an event, we define
P(E)=s∈E∑P(s).
That is the probability of an event is the sum of the probabilities of all the outcomes that comprise the event
This definition of probability also exhibits the properties of Finite probability mentioned earlier
P(Eˉ)=1−P(E)
P(E1∪E2)=P(E1)+P(E2)−P(E1∩E2)
and one further property
P(∪iEi)=∑iP(Ei)
for all pairwise disjoint events E4,E2,…
Examples
Let us have a die. that comes up 1 or 6 twice as often as all other numbers
Consider the following experiment A fair coin is tossed repeatedly until a Heads occurs (P(H)=P(T)=21)
The sample space consists of the number of tosses until we get an H
Clearly S={1,3,4,⋯}
This type of situation is called Repeated Bernoulli Trials
a Bernoulli Trial is an experiment with only 2 possible outcomes called success or failure
If p( success )=θ, then p( fail )=1−θ
Repeated Bernoulli Trails are when, we examine the outcome of independent trials the result on this flip of the coin doesn't depend on previous flips outcomes In this case the probability of a repeated trial is the product of the probability of each individual trial
For repeated flips of a fair coin
The probability that the first H occurs on the n th toss is 2n1 since this is the outcome
n−1 tails TTT⋯T⋅H← first head21⋅21⋯21⋅21=2n1
Define therefore the probability function
p(n)=2n1,n=1,2,3,⋯
Clearly 0≤p(n)≤1,∀n=1,2⋯so property i) is satisfied
Exercise: What is the probability that the first H occurs on,
an even numbered toss? , an odd numbered toss?
E1={ first H is on an even numbered toss }
E2={ first H is on an odd numbered toss }
Clearly E1=Eˉ2 so p(E1)=1−p(E2)
Now P(E1)P(E1)=31,P(E2)=32=n=1∑∞P(2n)=n=1∑∞22n1=n=1∑∞(41)n=41+421+431=41[1+41+421+431+⋯]=41⋅1−411=41⋅(43)1=31
(17)
\#2 on Exercise sheet for L19&20
What is the probability that a bit string of length 10 will
a) Start with 2 zeroes or end with 3 ones?
Here S={ bit strings of length 10}=Bit10
∣S∣=210E={ start with 00, er end with 111}E=E1∪E2E1={ start with 00}E2={ end with 111}∣E1∣=28∣E2∣=27∣E1∩E2∣=25∣E∣=28+27−25=352P(E)=210352=3211